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2002-P04904 - new septic system
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795 Ferndale Road North - 36-118-23-12-0014
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2002-P04904 - new septic system
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Last modified
8/22/2023 5:01:07 PM
Creation date
9/7/2016 1:26:27 PM
Metadata
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Template:
x Address Old
House Number
795
Street Name
Ferndale
Street Type
Road
Street Direction
North
Address
795 Ferndale Road North
Document Type
Septic
PIN
3611823120014
Supplemental fields
ProcessedPID
Updated
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'MOUND DESIGN WORK SHEET (For Flows u to 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows fn Gallons per Day <br /> num er o <br /> Estimated �o D D gpd (see figure A-1) bedrooms Class I Ciass II Class III Class IV <br /> or measured — x 1.5 (safety factor) _ — gpd 2 300 225 t 8o bo% <br /> 3 . 450 300 218 of the <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 900 525 332 Classl, <br /> �.—1 oc� � gallons (see figure G1) > >05o boo s7o ii, or u� <br /> 8 1200 675 408 columns, <br /> C. SOILS (ref2r t0 Stte 2Z7RIZl[it1011� C-1: Se ticTankCa acittes(in allons <br /> , Number of Minimum Liquid Liquid capacity with Liquid capaciry <br /> Bedrooms Ca aci arba e dis osal W��disposal& <br /> 1. Depth to restricting layer = a .0 feet P ry g g P liftinside <br /> 2. Depth of percolation tests = /.o feet 2orless �so �izs 15� <br /> 3. Texture c-�A� L--�� �„�, so�a �000 �soo z000 <br /> 5 or 6 ]500 2250 3� <br /> Percolation rate a�,�, mpi �,s o�9 z000 s000 <br /> 4. Soil loading rate . 4 � �nd/sqft(see figure D-33) <br /> 5. Percent land slope '� % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> (ec�c� gpd x 0.83 sqft/gpd = �4��C sqft <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x ; � gpd/sqft = ) v ft Mound LLR <br /> 3. Length of rock layer = area=width = <br /> '-1�� sqft (D1) = �c�_ft (D2) _ �ft < 120 M PI < � 2 <br /> E. ROCK VOLUME > 120 M P I < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> y� � sqft x 1 ft = ��cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> ��cuft = 27 cuyd/cuft = �_ cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> � cuyd x 1.4 ton/cuyd = a S� tons <br /> D-33: Absorptlon Width Sizing Table � <br /> F. SEWAGE ABSORPTION WIDTH Percoletion Rete Loading Rale <br /> in Minutes per Soil Texhue Gallons Absorption <br /> Inch per day per Ratio I <br /> MPI s uare foo� <br /> I <br /> . Faster Ihan 5 Coerse Sand 1.20 ].00 I <br /> Modium Smd '� <br /> Absorption width equals absorption ratio (See Figure D-33) Loamy Sand � <br /> times rock layer width (D2) �6� o Lo�, o.�o z o� ' <br /> 31 to 45 Silt Loam 0.50 2.a0 <br /> �l,�-���� x /�; ft = o-��-• '� ft 46�0 60 SandY Q.Y Lo �.45 2.6� <br /> Si l ty Cl ay Loam <br /> 61 to 120 Silty Cley 0.24 5.00 <br /> Sandy Cley <br /> la <br /> Slowerthan 120• <br /> •sy.��.�,eu��ee tor ihue.o��s m�.�n�aner or p�rroR,,.ae <br />
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